Question

The solution of the differential equation $\frac{dy}{dx}+x(x+y)=x^3(x+y{)}^3-1$ is $\frac{a}{(x+y{)}^b}=c(x^2+1)+k{e}^{x^2},$ then $a+b+c$ is equal to

Answer 1

$\frac{dy}{dx}+x(x+y)=x^3(x+y{)}^3-1$
$\Rightarrow \frac{1}{(x+y{)}^3}\left(\frac{dy}{dx}\right)+\frac{x}{(x+y{)}^2}-x^3=\frac{-1}{(x+y{)}^3}$
$\text{Let}t=\frac{1}{(x+y{)}^2}$
$\text{Then}\frac{dt}{dx}=\frac{-2}{(x+y{)}^3}(1+\frac{dy}{dx})$
$\text{So,}\frac{-1}{2}\frac{dt}{dx}+tx-x^3=0$
$\Rightarrow \frac{dt}{dx}-2tx=-2x^3$

$\text{I.F.}=e^{\int -2xdx}=e^{-x^2}$
Therefore, the solution is $t.e^{-x^2}=\int -2x^3.e^{-x^2}$
$\Rightarrow t.e^{-x^2}=e^{-x^2}(x^2+1)+C$
$\Rightarrow \frac{1}{(x+y{)}^2}=(x^2+1)+C{e}^{x^2}$
$\therefore a=1,b=2,c=1$
$\therefore a+b+c=4$